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# Two trains, X and Y, started simultaneously from opposite ends of a 10

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Joined: 02 Sep 2009
Posts: 50661
Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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Updated on: 01 Oct 2018, 01:04
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Question Stats:

75% (01:44) correct 25% (02:04) wrong based on 1311 sessions

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Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

Problem Solving
Question: 119
Category: Algebra Applied problems
Page: 77
Difficulty: 600

The Official Guide For GMAT® Quantitative Review, 2ND Edition

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Originally posted by Bunuel on 30 Nov 2010, 05:30.
Last edited by Bunuel on 01 Oct 2018, 01:04, edited 3 times in total.
Renamed the topic and edited the question.
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Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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27 Feb 2014, 06:06
5
61
Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.

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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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27 Feb 2014, 21:05
21
16
Train X's Average speed = 100/5 = 20 mph
Train Y's Average speed = 100/3 mph
Relative speed when trains travel toward each other = 20 + 100/3 = 160/3 mph.

Note that the 2 trains together would have traveled 100 miles at their meeting junction. Also, the time traveled for both trains would be the same at the meeting point.

Thus, Time = 100/(160/3) = 30/16 = 15/8 hours

Of the 100 miles, Distance covered by Train A = 20mph * 15/8 hours = 37.5 miles.

Ans is (A).
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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30 Nov 2010, 09:34
4
1
Train A speed = t1 = 20 mph
Train B speed= t2 = 100/3 mph
Combined speed(travelling towards each other) = 160/3 mph

When trains meet each other, they've covered 100 miles together.

Thus time = 100/(160/3) = 30/16 = 15/8 hours

out of 100 miles, distance covered by Train A = 20mph * 15/8 hours = 37.5 miles
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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30 Nov 2010, 09:59
19
8
ajit257 wrote:
Two trains X and Y started simultaneously from opposite ends of a 100 mile route and travelled toward each other on parallel tracks. Train X travelling at a constant rate completed the 100 mile trip in 5 hours. Train Y travelling at constant rate completed the 100 mile trip in 3 hours. How many miles had train X travelled when it met train Y ?

Can someone please explain the concept behind this type of problem ? All help appreciated.

The concept used in these questions is Relative Speed.

If two people walk in opposite directions (either towards each other or away from each other), their speed relative to each other is the sum of their speeds. e.g. If you are walking away from me at a speed of 2 miles/hr and I am walking away from you at a speed of 1 mile/hr, together we are creating a distance of 3 miles in 1 hr between us so our relative speed is 2 + 1 = 3 miles/hr
On the other hand, when two people walk in the same direction, their relative speed is the difference between their speeds.
e.g. if you are walking away from me at 1 mile/hr and I am walking towards you at 2 miles/hr, my speed relative to you is 2-1 = 1 mile/hr.

Time taken to meet = Total distance traveled/Relative speed

Speed of train X = 100/5 = 20 miles/hr
Speed of train Y = 100/3 miles/hr
Relative Speed = 20 + 100/3 = 160/3 miles/hr
Distance between them = 100 miles
Time taken to meet = 100/(160/3) hr = 15/8 hrs

In this time, train X would have traveled 20 * (15/8) = 37.5 miles

Faster Alternate Approach using Ratios :

Time taken by train X : Time taken by train Y = 5:3
Then, Speed of train X:Speed of train Y = 3:5
Since they start simultaneously, they travel for same time. So the ratio of their distance covered should be same as ratio of their speeds.
Distance covered by train X : Distance covered by train Y = 3:5
3/8 *100 = 37.5 miles (Distance covered by train X)
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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17 Feb 2011, 11:12
3
Two trains X and Y started simultaneously from opposite ends of a 100 mile route and travelled toward each other on parallel tracks. Train X travelling at a constant rate completed the 100 mile trip in 5 hours. Train Y travelling at constant rate completed the 100 mile trip in 3 hours. How many miles had train X travelled when it met train Y ?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

I solved as under
Time taken to meet = Total distance traveled/Relative speed

Speed of train X = 100/5 = 20 miles/hr
Speed of train Y = 100/3 miles/hr
Relative Speed = 20 + 100/3 = 160/3 miles/hr
Distance between them = 100 miles
Time taken to meet = 100/(160/3) hr = 15/8 hrs

In this time, train X would have traveled 20 * (15/8) = 37.5 miles
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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17 Feb 2011, 18:33
2
Baten80 wrote:
Two trains X and Y started simultaneously from opposite ends of a 100 mile route and travelled toward each other on parallel tracks. Train X travelling at a constant rate completed the 100 mile trip in 5 hours. Train Y travelling at constant rate completed the 100 mile trip in 3 hours. How many miles had train X travelled when it met train Y ?
(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

I solved as under
Time taken to meet = Total distance traveled/Relative speed

Speed of train X = 100/5 = 20 miles/hr
Speed of train Y = 100/3 miles/hr
Relative Speed = 20 + 100/3 = 160/3 miles/hr
Distance between them = 100 miles
Time taken to meet = 100/(160/3) hr = 15/8 hrs

In this time, train X would have traveled 20 * (15/8) = 37.5 miles

Another approach:
Time taken by train X to cover 100 miles : Time taken by train Y to cover 100 miles = 5:3
Therefore, Speed of X: Speed of Y = 3:5
So, when they meet, X would have covered (3/8)th of the total distance of 100 miles.
Distance covered by X = (3/8)*100 = 37.5 miles
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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27 Feb 2014, 23:31
5
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Explaination:

Let the point where they meet is x kms from Point X

So it is (100-x) from point Y

Ratio of distance = Ratio of time taken

x/(100-x) = 20 / (100/3)

Solving the above equation, we get x = 37.5
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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29 May 2015, 20:40
1
As per the OG explanations:

To solve this problem, use the formula distance = rate x time and its two equivalent forms rate = distance and time = distance. Train X time rate traveled 100 miles in 5 hours so ts rate was 100/5 = 20 miles per hour. Train Y traveled 100 miles in 3 hours so its rate was 1003 miles per hour. If t represents the number of hours the trains took to meet, then when the trains met, Train X had traveled a distance of 20t miles and Train Y had traveled a distance of 100/3 t miles.

How can "t" represent the time for both the trains? Because they have different rates, doesn't that mean each will take different time to meet?

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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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30 May 2015, 14:55
Hi Pretz,

In this question, the variable "T" represents "amount of time that each train was moving" - since both trains started moving SIMULTANEOUSLY, the "T" can be used in both calculations ("T" does NOT represent the "time of day"). Since the trains are moving at DIFFERENT RATES, the DISTANCE that each train will travel will be different.

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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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31 May 2015, 23:43
2
Pretz wrote:
As per the OG explanations:

To solve this problem, use the formula distance = rate x time and its two equivalent forms rate = distance and time = distance. Train X time rate traveled 100 miles in 5 hours so ts rate was 100/5 = 20 miles per hour. Train Y traveled 100 miles in 3 hours so its rate was 1003 miles per hour. If t represents the number of hours the trains took to meet, then when the trains met, Train X had traveled a distance of 20t miles and Train Y had traveled a distance of 100/3 t miles.

How can "t" represent the time for both the trains? Because they have different rates, doesn't that mean each will take different time to meet?

Think of it this way:

You are at your home and your friend is at his home. You both decide to meet. You leave your respective homes at exactly 12:00 and then travel toward each other's homes at your own speeds. You meet i.e. reach the same point, at say, 12:20. Have you traveled for the same amount of time? Sure. You both have traveled for exactly 20 mins. You traveled at your own speeds: say you are very fast and your friend is very slow. So how does this impact the entire equation? You would have covered much more distance than your friend in the same 20 mins. So higher speed will lead to more distance covered but the time for which the two of you would have traveled would be the same. Similarly, since the trains start at the same time, when they meet, the time elapsed would be the same. They will cover different distances due to their different speeds.
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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18 Jun 2015, 03:15
2
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

Problem Solving
Question: 119
Category: Algebra Applied problems
Page: 77
Difficulty: 600

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Is this approach recommendable?:

speed x = 100/5 = 20mph
speed y = 100/3 = 33,3mph

after 1h: X = 20miles, Y = 33,3 miles
after 2h: X= 40 miles, Y = 66,6 miles --> together 106,6 miles ... total dist = 100 miles..they have met in this time frame!
as 106,6 miles, X traveled not 40 but slightly lower --> 37,5 --> Answer A
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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18 Jun 2015, 03:45
LaxAvenger wrote:
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

Problem Solving
Question: 119
Category: Algebra Applied problems
Page: 77
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

Is this approach recommendable?:

speed x = 100/5 = 20mph
speed y = 100/3 = 33,3mph

after 1h: X = 20miles, Y = 33,3 miles
after 2h: X= 40 miles, Y = 66,6 miles --> together 106,6 miles ... total dist = 100 miles..they have met in this time frame!
as 106,6 miles, X traveled not 40 but slightly lower --> 37,5 --> Answer A

Yes, It's absolutely correct approximation with the given options.

Instead you could have done another thing here

Distance of X/Distance of Y = speed of X/speed of Y

a/(100-a) = 20/33.3

a/(100-a) = 6/10

10a = 600-6a

16a = 600

a = 37.5
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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01 Feb 2016, 20:17
3
No calculations approach.
By the end of hr1 A would’ve traveled 20 miles and B 33 1/3 miles. So they have not met yet
By the end of hr2 A would’ve traveled 40 miles and B 66 2/3 miles. So they have already crossed. Thus A traveled less than 40 miles when it first met B. The only choice less than 40 miles is A
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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07 Feb 2017, 04:26
Bunuel isn't the ratio of X to Y suppose to be 5 : 3 and not 3 to 5?
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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07 Feb 2017, 05:05
2
iliavko wrote:
Bunuel isn't the ratio of X to Y suppose to be 5 : 3 and not 3 to 5?

Train X, travelling at a constant rate, completed the 100-mile trip in 5 hours --> rate of X = (distance)/(time) = 100/5 =20 miles per hour;
Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours --> rate of Y = (distance)/(time) = 100/3 miles per hour;

(rate of X)/(rate of Y) = 20/(100/3) = 20*3/100 = 3/5.
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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09 Feb 2017, 16:29
2
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

We are given that train X completed the the 100-mile trip in 5 hours, and that train Y completed the 100-mile trip in 3 hours.

Since rate = distance/time, the rate of train X is 100/5 = 20 mph and the rate of train Y is 100/3 mph.

Since the trains left at the same time, we can let the time of each train = t.

We need to determine the distance traveled by train X when it met train Y. Since the two trains are “converging” we can use the formula:

distance of train X + distance of train Y = total distance

20t + (100/3)t = 100

Multiplying the entire equation by 3, we have:

60t + 100t = 300

160t = 300

t = 300/160 = 30/16 = 15/8.

Thus, train X and Y met each other after 15/8 hours.

Since distance = rate x time, the distance traveled by train X when it met train Y was:

15/8 x 20 = 300/8 = 75/2 = 37.5 miles.

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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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30 May 2017, 16:17
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

1. They would have met in 100/(100/5+100/3) hrs =15/8 hrs
2. In that time X would have traveled, Time *speed =15/8 *20=37.5
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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25 Jun 2017, 14:31

I like to sketch out the problem and many times it is a fast method.

As you can see train A meets train B somewhere between the 20 and 40th mile. Looking at the answer choices, it is pretty obvious that A is the answer.
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Re: Two trains, X and Y, started simultaneously from opposite ends of a 10  [#permalink]

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07 Dec 2017, 09:24
Top Contributor
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?

(A) 37.5
(B) 40.0
(C) 60.0
(D) 62.5
(E) 77.5

Train X completed the 100-mile trip in 5 hours
Speed = distance/time
= 100/5
= 20 mph

Train Y completed the 100-mile trip in 3 hours
Speed = distance/time
= 100/3
33 mph (This approximation is close enough. You'll see why shortly)

How many miles had Train X traveled when it met Train Y?

When the two trains meet, each train will have been traveling for the same amount of time
So, we can write: Train X's travel time = Train Y's travel time

time = distance/speed
We know each train's speed, but not the distance traveled (when they meet). So, let's assign some variables.

Let d = the distance train X travels
So, 100-d = the distance train Y travels (since their COMBINED travel distance must add to 100 miles)

We can now turn our word equation into an algebraic equation.
We get: d/20 = (100 - d)/33
Cross multiply to get: (33)(d) = (20)(100 - d)
Expand: 33d = 2000 - 20d
Add 20d to both sides: 53d = 2000
So, d = 2000/53

IMPORTANT: Before you start performing any long division, first notice that 2000/50 = 40
Since the denominator is greater than 50, we can conclude that 2000/53 is LESS THAN 40
Since only one answer choice is less than 40, the correct answer must be A

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